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3 years ago

12

james swims at a speed of 2 m/s when she notices a shark is behind her and accelerates so that she is skimming at 4 m/s. if her

average acceleration is 1.2 m/s how long did it take for james to speed up

1 answer:

Veronika [31]3 years ago

3 0

Answer:

Time= t= 1.67 secs

Explanation:

Time= t=?= 1.67 secs

Initial Velocity= Vi = 2m/s

Final Velocity= Vf = 4m/s

Average Acceleration= av= Vf-Vi/t

Putting the values : 1.2m/s²= 4-2/t

Rearranging: t= 2/1.2= 1.67 secs

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A fighter jet accelerates from 10 m/s to 75 m/s in 7.0 s. The acceleration of the jet is

liraira [26]

Acceleration = final velocity - inital / time
a = 75-10 / 7
a = 65 / 7
a = 9.29 m/s^2

4 0

3 years ago

If an equation is dimensionally correct is thar equation a right equation<br>

Nady [450]

If an equation is dimensionally correct, it does not mean that the equation must be true. On the other hand, when the equation is dimensionally correct, the equation cannot be true. Dimensional analysis is a technique used to check whether a relationship is correct

8 0

3 years ago

Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on

Natasha2012 [34]

Answer:

$8.8\times 10^{-6} W/m^2$

Explanation:

We are given that

Wavelength, $\lambda=571 nm=571\times 10^{-9} m$

$1 nm=10^{-9} m$

R=75 cm= $\frac{75}{100}=0.75 m$

1 m=100 cm

$d=0.640 mm=0.64\times 10^{-3} m$

$1 mm=10^{-3} m$

$a=0.434 mm=0.434\times 10^{-3} m$

$y=0.830 mm=0.830\times 10^{-3} m$

$I_0=5.00\times 10^{-4}W/m^2$

$tan\theta=\frac{y}{R}$

$\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}$

$\theta=1.1\times 10^{-3}rad$

$tan\theta\approx \theta$

Because $\theta$ is small.

$\phi=\frac{2\pi dsin\theta}{\lambda}$

$\sin\theta\approx \theta$ ,

Therefore

$\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}$

$\phi=7.74 rad$

$\beta=\frac{2\pi a\theta}{\lambda}$

$\beta=5.3 rad$

$I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2$

$I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2$

$I=8.8\times 10^{-6} W/m^2$

6 0

3 years ago

A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the block in water?

12345 [234]

   Density = (mass) / (volume)

      4,000 kg/m³ = (mass) / (0.09 m³)

Multiply each side
by 0.09 m³ :           (4,000 kg/m³) x (0.09 m³) = mass

   mass = 360 kg .

Force of gravity = (mass) x (acceleration of gravity)

   = (360 kg) x (9.8 m/s²)

   = (360 x 9.8) kg-m/s²

   = 3,528 newtons .

That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).

Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.

The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.

So while it's in the water, the block seems to weigh

   (3,528 - 882) = 2,646 newtons (about 595.2 pounds) .

But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.

5 0

4 years ago

Read 2 more answers

Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements

masya89 [10]

Answer:

When the starting and ending points are the same, the total work is zero.

Explanation:

option ( D )correct

A force is said to be conservative when the work done by the force in moving a particle from a point A to a point B is independent of the path followed between A and B and is the same for all the paths. The work done depends only on the particles initial and final positions. And when the initial and final position in conservative field are same the work done is said to be zero.

8 0

3 years ago